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3 years ago

Suppose you held two magnets a short distance apart,then let go. what would happen?

Chemistry

2 answers:

Murljashka [212]3 years ago

4 0

The two magnets ought to join together given that at least, if not both, have a strong enough magnetic force to do so.

Assoli18 [71]3 years ago

4 0

Answer:

There are two possibilities that would occur:

The magnets come together and stick together.
The magnets are further apart than the distance at which they were held.

Explanation:

The magnets have two poles: positive and negative, before which it is recognized that <u>if two magnets with the same pole are close to each other and are released, these magnets will repel each other</u>, this is because the same pole repels, regardless of whether it is negative or positive, on the contrary, <u>if the two magnets are close and have opposite poles directed towards each other, these magnets will attract</u>, even with this there is usually a saying that "the opposite poles attract," although it is usually use in couple relationships, this saying may allow you to remember this rule.

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3 years ago

Which statement is always true about a reversible chemical reaction?

Alecsey [184]

<h2>Answer:</h2>

Option (B):

The products can form reactants, and the reactants can form products.

<h3>Explanation:</h3><h3>Reversible reaction</h3>

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aA + bB ⇄ cC + dD

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3 years ago

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3 years ago

A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at

Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

$T_{final}=80.30^oC$

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0

2 years ago